GPS : SC3-d , Students will relate light emission and the movement of the electrons to element identification.
Authors : Michala Metcalf and Olenkeya Irvin Abstract : Perform flame tests of metal cations to observe their characteristic colors and match the colors observed to an appropriate wavelength of visible light.
Method : Equipment :
6 Wooden Splints (pre-soaked in distilled water)
Wash Bottle
Bunsen Burner
250 ml beaker
400 ml beaker
Distilled water
Solid Salts
LiCl
NaCl
KCl
CuCl2
BaCl2
CaCl2
Procedure :
Gather all equipment and chemicals.
Set up lab space.
Pour Distilled water into the 250 ml beaker and pre-soak the 6 wooden splints.
Light your Bunsen burner, then adjust the flame to a moderate temperature (no yellow in the flame)
Begin by dipping the end of a water soaked splint into one of the solid compounds. The solid crystals should coat the end of the splint.
Hold the coated end of the splint in the central portion, but just to the side of it, of the Bunsen burner flame.
Record the dominant color observed and its associated wavelength.
Repeat instruction 5-7 for the other 5 solids.
Dominant Color
Approximate Wavelength(in NM)*
Compound
Red
701
LiCl
Orange
609
CaCl2
Orange-Yellow
597
NaCl2
Yellow-Green
577
BaCl2
Green-Blue
492
CuCl2
Violet
423
KCl
Formulas :E = H x V (energy = plank's constant x frequency) H = 6.626 x 10 -34 V = C /λ ( frequency = velocity / wavelength) C = 3.00 x 10 10 nm = 10 -9m = 10 -7 m = 10 -5 cm
Calculations : LiCl - V = (3.00 x 10 10) / (7.01 x 10 -5) = .42796 x 10 15 = 4.28 x 10 14
E = (6.626 x 10 -34) x (4.28 x 10 14) = 28.35928 x 10 -20 =2.84 x 10 -19
CaCl2 - V = (3.00 x 10 10) / ( 6.09 x 10 -5) = .4926108374 x 10 15 = 4.93 x 10 14
E= (6.626 x 10 -34) x (4.93 x 10 14) =32.66618 x 10 -20 = 3.27 x 10 -19
NaCl2 - V = (3.00 x 10 10) / (5.97 x 10-5) =.5025125628 x 10 15 = 5.03 x 10 14
E = (6.626 x 10 -34) x (5.03 x 10 14) = 33.32878 x 10 -20 = 3.33 x 10 -19
BaCl2 - V = (3.00 x 10 10) / (5.77 x 10 15) = .5199306759 x 10 15 = 5.20 x 19 14
E = (6.626 x 10 -34) x (5.20 x 19 14) = 34.4552 x 10 -20 = 3.45 x 10 -19
CuCl - V= (3.00 x 10 10) / (4.92 x 10 -5) = .6097560976 x 10 15 = 6.10 x 10 14
E = (6.626 x 10 -34) x (6.10 x 10 14) = 40.4186 x 10 -20 = 4.04 x 10 -19
KCl - V = (3.00 x 10 10) / (4.23 x 10 -5) = 12.69 x 10 15 = 1.27 x 10 14
E = (6.626 x 10 -34) x (1.27 x 10 14) = 8.41502 x 10 -20 = 8.42 x 10 -20
Error Analysis : There isn't a theoretical value to calculate the error .
Conclusion : Doing this lab let us find out the color and wave length the chemical produced. Each substances create a different level of energy produced. That means that every chemical will produce a different color in the flame test. The reason behind that would be because of the different energy levels. When an electron jump from one level to another and jump back; that shows the production of light.
GPS : SC3-d , Students will relate light emission and the movement of the electrons to element identification.
Authors : Michala Metcalf and Olenkeya Irvin
Abstract : Perform flame tests of metal cations to observe their characteristic colors and match the colors observed to an appropriate wavelength of visible light.
Method :
Equipment :
- 6 Wooden Splints (pre-soaked in distilled water)
- Wash Bottle
- Bunsen Burner
- 250 ml beaker
- 400 ml beaker
- Distilled water
Solid SaltsProcedure :
Formulas : E = H x V (energy = plank's constant x frequency) H = 6.626 x 10 -34
V = C / λ ( frequency = velocity / wavelength) C = 3.00 x 10 10
nm = 10 -9m = 10 -7 m = 10 -5 cm
Calculations :
LiCl - V = (3.00 x 10 10) / (7.01 x 10 -5)
= .42796 x 10 15
= 4.28 x 10 14
E = (6.626 x 10 -34) x (4.28 x 10 14)
= 28.35928 x 10 -20
=2.84 x 10 -19
CaCl2 - V = (3.00 x 10 10) / ( 6.09 x 10 -5)
= .4926108374 x 10 15
= 4.93 x 10 14
E= (6.626 x 10 -34) x (4.93 x 10 14)
=32.66618 x 10 -20
= 3.27 x 10 -19
NaCl2 - V = (3.00 x 10 10) / (5.97 x 10-5)
=.5025125628 x 10 15
= 5.03 x 10 14
E = (6.626 x 10 -34) x (5.03 x 10 14)
= 33.32878 x 10 -20
= 3.33 x 10 -19
BaCl2 - V = (3.00 x 10 10) / (5.77 x 10 15)
= .5199306759 x 10 15
= 5.20 x 19 14
E = (6.626 x 10 -34) x (5.20 x 19 14)
= 34.4552 x 10 -20
= 3.45 x 10 -19
CuCl - V= (3.00 x 10 10) / (4.92 x 10 -5)
= .6097560976 x 10 15
= 6.10 x 10 14
E = (6.626 x 10 -34) x (6.10 x 10 14)
= 40.4186 x 10 -20
= 4.04 x 10 -19
KCl - V = (3.00 x 10 10) / (4.23 x 10 -5)
= 12.69 x 10 15
= 1.27 x 10 14
E = (6.626 x 10 -34) x (1.27 x 10 14)
= 8.41502 x 10 -20
= 8.42 x 10 -20
Error Analysis : There isn't a theoretical value to calculate the error .
Conclusion : Doing this lab let us find out the color and wave length the chemical produced. Each substances create a different level of energy produced. That means that every chemical will produce a different color in the flame test. The reason behind that would be because of the different energy levels. When an electron jump from one level to another and jump back; that shows the production of light.
Safety Precautions :